An empty 4.00-L steel vessel is filled with 1.00 atm of CH 4 (g) and 4.00 atm of O 2 (g) at 300 °C. A spark causes the CH 4 to burn completely, according to the equation CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) ΔH° = -802 kJ (a) What mass of CO 2 (g) is produced in the reaction?

Hey everyone. So here we give them the following reaction that is unbalanced. and at 300 Kelvin, A 2.5 L reaction chamber is composed of chlorine gas and florian gas With partial pressures of 472 mm of mercury and 806 2 mm of Mercury were asked to determine the limiting reagent. And also what is the theoretical yield of clothing traveler rides. We need to first balance the reaction. FCL two. Yes plus F two. Yes. And it's Gonna Yell Clf three. Yes, we have chlorine and chlorine on both sides. We have to climbing over here to flooring over here, one flooring over here and free flooring over here. Since we have two climbing on the acting side and one on the product side, we can also have R2 over here. It too Put a two in front of clf three And I have six florin over here. I can put it in front of F2 over here to get six flooring over here as well. And now this is balanced. And now we need to use the ideal gas law to determine the number of moles of chlorine gas and fluorine gas. It's going to be PV because N. R. T. Our pressure 472 millimeters of mercury, The volume is 2.5 leaders and we look for the number of moles are constant 0.08-1 atmosphere. Bye bye. Malls times kelvin And the temperature is salman Now we need to convert our pressure from millimeters of mercury atmosphere. We have 472 millimeters of mercury And we have 760 millimeters of mercury in one atmosphere. So this will give us 0.6-1 atmosphere. So now if we plug in the bios, we get 0.6-1 atmosphere Times 2.5 leaders eager to end Time, 21 later. Some atmosphere. What about moles kelvin Times salman. This will give us 1.55 two side eagles, 24 .71-1 comes in. If you divide both sides 5 24 7121, we're gonna get in Equals 0. 28 moles. And this is using chlorine gas. So if we use flurry and gas, we're gonna have the pressure. It was a 162 millimeters of mercury And the volume is 2.5 Leaders for the number of moles, Our constant is 0.08 21 later. Sounds atmosphere about about moles, name's Calvin And the temperature standing on one Calvin. And if you convert the pressure millimeters of mercury to atmosphere, We're gonna get 862 millimeters of mercury. We have 760 millimeters of mercury and one atmosphere And this will give us 1. atmosphere. Now, if you plug into the place you're going to get 1.134 atmosphere Times 2.5 leaders. it goes in From 0. 21 later. Some atmosphere about about moles. I'm kelvin Times 301 Kelvin And this gives us 2.835. It was an I'm 24 so feet about both sides 24.71 - one. We're gonna get in Even 0. 47 most. And now we need to use the malls of each gas to determine the mass of chlorine to fluoride. 0.062 Malls of Cl two. And in one bowl cl two We have two moles of CLF three And in one mole LCLF three like the Mueller mass. And this is 35. plus three times 18 .998. Don't you give us money too? 0.45. It's gonna be 11.46 38. And if you have 0.11 or seven Balls of F two, We have three malls of F two people to to malls of CL. F three on one mob of CLF three We have 0.45 grams of CLF three. And this give a 7.07 g CLF three. So our limiting reagent and the flooring gas because this produced a lesser amount of chlorine fluoride. And the theoretical yield 27.07 g of clothing traveler ride. Thanks for watching my video, and I hope it was helpful.

Ch.10 - Gases: Their Properties & Behavior

All textbooks

McMurry 8th Edition

Ch.10 - Gases: Their Properties & Behavior

Problem 145a

Chapter 10, Problem 145a

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ
(a) What mass of CO₂(g) is produced in the reaction?

Verified step by step guidance

Identify the balanced chemical equation for the reaction: \( \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \).

Calculate the initial moles of \( \text{CH}_4 \) and \( \text{O}_2 \) using the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.

Determine the limiting reactant by comparing the mole ratio of \( \text{CH}_4 \) to \( \text{O}_2 \) with the stoichiometric ratio from the balanced equation.

Use the stoichiometry of the balanced equation to calculate the moles of \( \text{CO}_2 \) produced, based on the limiting reactant.

Convert the moles of \( \text{CO}_2 \) to mass using the molar mass of \( \text{CO}_2 \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. In this case, understanding the stoichiometric coefficients from the combustion reaction of methane (CH4) is essential to determine how much carbon dioxide (CO2) is produced.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is crucial for calculating the number of moles of gases involved in the reaction, especially since the problem provides initial pressures and a specific volume. By applying the Ideal Gas Law, we can convert the given pressures of CH4 and O2 into moles, which are necessary for stoichiometric calculations.

Enthalpy Change (ΔH)

Enthalpy change (ΔH) represents the heat absorbed or released during a chemical reaction at constant pressure. In this combustion reaction, the negative ΔH value indicates that the reaction is exothermic, meaning it releases heat. Understanding the enthalpy change helps in assessing the energy dynamics of the reaction, although it is not directly needed to calculate the mass of CO2 produced, it provides context for the reaction's energy profile.

Recommended video:

Guided course

02:34

Enthalpy of Formation

Related Practice

Textbook Question

A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (a) Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 K.

Textbook Question

A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (b) Now make a more accurate calculation of the pressure inside the container. Take into account the volume of solid CaO (density = 3.34 g/mL) in the container, and use the van der Waals equation to calculate the pressure. The van der Waals constants for CO₂(g) are a = 3.59 (L²-atm)/mol² and b = 0.0427 L/mol.

Textbook Question

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ

(b) What is the final temperature inside the vessel after combustion, assuming that the steel vessel has a mass of 14.500 kg, the mixture of gases has an average molar heat capacity of 21 J/(mol·°C), and the heat capacity of steel is 0.449 J/(g·°C)?

Textbook Question

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ

An empty 4.00-L steel vessel is filled with 1.00 atm of CH4(g) and 4.00 atm of O2(g) at 300 °C. A spark causes the CH4 to burn completely, according to the equationCH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH° = -802 kJ(a) What mass of CO2(g) is produced in the reaction?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Ideal Gas Law

Enthalpy Change (ΔH)

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ
(a) What mass of CO₂(g) is produced in the reaction?